A prime-power cubic binomial supercongruence for primes congruent to 3 modulo 4

From papers

Let p>3p>3 be a prime with p3(mod4)p\equiv3\pmod4, and let rr be a positive integer. The supercongruence.

k=0pr1164kj=0k(2jj)3(2k2jkj)30(modp4).\sum_{k=0}^{p^r-1}\frac1{64^k}\sum_{j=0}^{k}{2j\choose j}^3{2k-2j\choose k-j}^3\equiv0\pmod{p^4}.

This is described as a generalization of an earlier case distinction; the stated modulus is independent of rr, and the source gives no proof.

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Sources & referencesView supporting material

Primary source

Victor J. W. Guo and Long Li, “q-Supercongruences from squares of basic hypergeometric series”, arXiv:2112.12076 (2021).

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